Consider the circle C : x² + y² = 4 and the parabola P : y² = 8x. If the set of all values of α, for

Q: Wait, why do we use $T = S_1$ for the chord equation?

That is a fantastic question! In coordinate geometry, $T = S_1$ is a standard, elegant formula for the chord of a conic section (like a circle or parabola) when the midpoint $(x_1, y_1)$ is known. It saves us from the tedious process of finding the slope, writing the line equation, and then finding the intersection points. It directly gives us the equation of the chord in terms of its midpoint, which is exactly what we need here.

Q: Why is the condition $x_1^2 + y_1^2 < 4$ necessary?

Think about the definition of a chord. A chord is a line segment whose endpoints lie on the circle. If the midpoint of a line segment lies outside the circle, the segment cannot possibly be a chord of that circle! Therefore, for the segment to be a valid chord, its midpoint must lie strictly inside the circle. This is a classic 'hidden' constraint in JEE problems that separates the top rankers from the rest.

Q: Why do we need $x_1 = \alpha - 8$ to give two distinct points?

The problem asks for three distinct chords. We found that $x_1 = 0$ gives us one chord (the x-axis). To get three chords in total, the other root, $x_1 = \alpha - 8$, must provide two distinct, valid midpoints. This means $x_1$ must be non-zero and $y_1^2$ must be positive, which leads us to the condition $\alpha > 8$. It is all about ensuring the geometry matches the algebraic requirements!

Question

Consider the circle $C : x^2 + y^2 = 4$ and the parabola $P : y^2 = 8x$. If the set of all values of $\alpha$, for which three chords of the circle $C$ on three distinct lines passing through the point $(\alpha, 0)$ are bisected by the parabola $P$ is the interval $(p, q)$, then $(2q - p)^2$ is equal to ________

Sigma Percentile · Accepted Answer

### Analyzing the Setup Welcome, fellow traveler on the path to JEE mastery! Today, we are going to dissect a problem that perfectly encapsulates the beauty of coordinate geometry. It is not just about crunching numbers; it is about visualizing the dance between a circle and a parabola. Imagine you are standing on the Cartesian plane. You see the circle $C: x^2 + y^2 = 4$, a perfect, symmetric shape centered at the origin with a radius of $2$. Beside it, the parabola $P: y^2 = 8x$ opens its arms to the right. We are given a point $A(\alpha, 0)$ on the x-axis, and we are tasked with finding the range of $\alpha$ such that three distinct chords of the circle, all passing through $A$, are bisected by this parabola. ### The Magic of $T = S_1$ When we talk about a chord of a circle with a known midpoint $M(x_1, y_1)$, we do not need to struggle with slopes or complex intersections. We have a secret weapon: the formula $T = S_1$. For our circle $C: x^2 + y^2 - 4 = 0$, the equation of the chord with midpoint $(x_1, y_1)$ is given by: $$xx_1 + yy_1 - 4 = x_1^2 + y_1^2 - 4$$ Simplifying this, we get: $$xx_1 + yy_1 = x_1^2 + y_1^2$$ This equation is the heartbeat of our solution. It defines the line segment that is perfectly bisected at $(x_1, y_1)$. ### The Parabolic Bridge Our chord must pass through the point $A(\alpha, 0)$. This is our anchor. By substituting $x = \alpha$ and $y = 0$ into our chord equation, we get: $$\alpha x_1 = x_1^2 + y_1^2$$ The problem states that these chords are bisected by the parabola $P: y^2 = 8x$. This means the midpoint $M(x_1, y_1)$ must lie on the parabola, so we can replace $y_1^2$ with $8x_1$. Substituting this into our equation, we get: $$\alpha x_1 = x_1^2 + 8x_1$$ Rearranging this yields: $$x_1^2 + (8 - \alpha)x_1 = 0 \Rightarrow x_1(x_1 + 8 - \alpha) = 0$$ This gives us two potential x-coordinates for our midpoints: $x_1 = 0$ and $x_1 = \alpha - 8$. ### The Reality Check Now, we must be careful. We need three distinct chords. The case $x_1 = 0$ gives us the midpoint $(0,0)$, which corresponds to the x-axis as our first chord. For the other root, $x_1 = \alpha - 8$, to provide two more distinct chords, we need $y_1^2 > 0$. Since $y_1^2 = 8x_1$, this implies $8(\alpha - 8) > 0$, or $\alpha > 8$. There is a final, crucial constraint: the midpoint must lie inside the circle. If the midpoint is outside, the chord does not exist! Thus, we must satisfy $x_1^2 + y_1^2 < 4$. Substituting $x_1 = \alpha - 8$ and $y_1^2 = 8(\alpha - 8)$, we get the inequality: $$(\alpha - 8)^2 + 8(\alpha - 8) < 4$$ Let $t = \alpha - 8$. The inequality becomes $t^2 + 8t - 4 < 0$. Solving the quadratic $t^2 + 8t - 4 = 0$ gives $t = -4 \pm 2\sqrt{5}$. Thus, $-4 - 2\sqrt{5} < \alpha - 8 < -4 + 2\sqrt{5}$, which simplifies to: $$4 - 2\sqrt{5} < \alpha < 4 + 2\sqrt{5}$$ ### Final Calculation We have two conditions: $\alpha > 8$ and $4 - 2\sqrt{5} < \alpha < 4 + 2\sqrt{5}$. Since $4 + 2\sqrt{5} \approx 8.47$, the intersection is: $$8 < \alpha < 4 + 2\sqrt{5}$$ Here, $p = 8$ and $q = 4 + 2\sqrt{5}$. We need to calculate $(2q - p)^2$. Substituting our values: $$2q - p = 2(4 + 2\sqrt{5}) - 8 = 8 + 4\sqrt{5} - 8 = 4\sqrt{5}$$ Squaring this, we get: $$(4\sqrt{5})^2 = 16 imes 5 = \mathbf{80}$$ And there it is! A beautiful, logical path to the answer. Never fear the complexity; just break it down, step by step, and the geometry will reveal its secrets.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e G eo m e t r y$

$E q u a t i o n o f a B i sec t e d C h or d$

$P a ss in g T h r o ug h (α, 0)$

$M i d p o in t o n t h e P a r ab o l a$

$S o l v in g f or x_{1}$

$C o n d i t i o n f or T h r ee D i s t in c t C h or d s$

$C o n s t r ain t : M i d p o in t I n s i d e t h e C i r c l e$

$S e tt in g u p t h e I n e q u a l i t y$

$S o l v in g t h e Q u a d r a t i c I n e q u a l i t y$

$R e v er t in g t o α$

$F in d in g t h e F ina l R an g e (p, q)$

$C a l c u l a t in g (2 q - p)^{2}$

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Magic of $T = S_{1}$

The Parabolic Bridge

The Reality Check

Final Calculation

Similar Questions

A circle of radius 2 unit passes through the vertex and the focus of the parabola $y^{2} = 2 x$ and touches the parabola $y = (x - \frac{1}{4})^{2} + α$ , where $α > 0$ . Then $(4 α - 8)^{2}$ is equal to ____.

Let $S$ be the focus of the parabola $y^{2} = 8 x$ and let $P Q$ be the common chord of the circle $x^{2} + y^{2} - 2 x - 4 y = 0$ and the given parabola. The area of the triangle $P QS$ is

Through the vertex $O$ of parabola $y^{2} = 4 x$ , chords $O P$ and $O Q$ are drawn at right angles to one another. Show that for all positions of $P$ , $P Q$ cuts the axis of the parabola at a fixed point. Also find the locus of the middle point of $P Q$ .

Comprehension Passage

The ratio of the areas of the triangles $P QS$ and $P QR$ is

The radius of the circumcircle of the triangle $P R S$ is

The radius of the incircle of the triangle $P QR$ is

If the points of intersection of two distinct conics $x^{2} + y^{2} = 4 b$ and $\frac{x ^{2}}{16} + \frac{y ^{2}}{b ^{2}} = 1$ lie on the curve $y^{2} = 3 x^{2}$ , then $33$ times the area of the rectangle formed by the intersection points is ______

Let $P$ and $Q$ be distinct points on the parabola $y^{2} = 2 x$ such that a circle with $P Q$ as diameter passes through the vertex $O$ of the parabola. If $P$ lies in the first quadrant and the area of the triangle $Δ O P Q$ is $32$ , then which of the following is (are) the coordinates of $P$ ?

The common tangents to the circle $x^{2} + y^{2} = 2$ and the parabola $y^{2} = 8 x$ touch the circle at the points $P, Q$ and the parabola at the points $R, S$ . Then the area of the quadrilateral $P QR S$ is

Let the length of the focal chord $P Q$ of the parabola $y^{2} = 12 x$ be 15 units. If the distance of $P Q$ from the origin is $p$ , then $10 p^{2}$ is equal to _______

The length of the chord of the parabola $x^{2} = 4 y$ having equation $x - 2 y + 42 = 0$ is :

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Geometry

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of a Bisected Chord

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Passing Through (α,0)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Midpoint on the Parabola

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for x1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Three Distinct Chords

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Constraint: Midpoint Inside the Circle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Inequality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving the Quadratic Inequality

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reverting to α

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Final Range (p,q)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating (2q−p)2